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I have the following definition:

Let ($X$,$\mathcal{T}$) and ($X'$, $\mathcal{T'}$) be topological spaces. A surjection $q: X \longrightarrow X'$ is a quotient mapping if $$U'\in \mathcal{T'} \Longleftrightarrow q^{-1}\left( U'\right) \in \mathcal{T} \quad \text{i.e. if } \mathcal{T'}=\{ U' \subset X' : q^{-1}\left( U' \right) \in \mathcal{T} \}$$

and the properties:

  1. $q$ is a bijective quotient mapping $\Leftrightarrow$ $q$ is a homeomorphism
  2. In general, $q$ quotient $\not \Rightarrow q$ open. If $U \in \mathcal{T}, q(U)\subset X'$ is open if $q^{-1}\left( q\left( U \right) \right) \in \mathcal{T}$ but not in general.

I could not find an example of quotient mapping for which $q^{-1}\left( q\left( U \right) \right)$ is not open. I would understand the idea better if you could show me one.

Martin Sleziak
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Luc M
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6 Answers6

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Consider $\mathbb{R}$ with the standard topology. On $\mathbb{R}$, consider the equivalence relation

$$x\sim y \iff (x = y \lor \{x,y\} \subset \mathbb{Z}),$$

and let $(X',\mathcal{T}')$ the quotient space $\mathbb{R}/{\sim}$. By definition, $\pi \colon \mathbb{R}\to X';\; x \mapsto [x]_\sim$ is a quotient map, but that map is not open:

If $U \subset\mathbb{R}$ is an open set containing an integer, then $\pi^{-1}(\pi(U)) = U\cup \mathbb{Z}$ is in general not open.

Alex Ortiz
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Daniel Fischer
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Here is a "visual" example:

Wrap the unit interval $[0,1]$ around the unit circle $S^1$ by identifying $0$ and $1$. The interval $[0,1/2) $ is open in $[0,1]$, but its image in $S^1$ is not open.

Ken
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  • Why the image isn't open? Can you explain? – SoG Jun 22 '23 at 16:22
  • @Sourav Ghosh Because its preimage in $[0,1]$ is not open. (Recall that if $p:X\to Y$ is a quotient map of topological spaces, then a subset of $Y$ is open if and only if its preimage in $X$ is open.) – Ken Jun 23 '23 at 02:31
  • OK. $f^{-1}f{[0, \frac{1}{2})}=[0, \frac{1}{2}) \cup{1}$ – SoG Jun 23 '23 at 05:08
  • Can you please explain how is it a quotient mapping? Thank you in advance. – Rajdeep Mar 10 '24 at 12:34
  • @Rajdeep It is a closed map by https://math.stackexchange.com/questions/1728663/continuous-map-on-compact-hausdorff-spaces-is-closed, and a continuous, closed, surjective map is a quotient map https://math.stackexchange.com/questions/3981375/a-surjective-continuous-and-closed-map-is-a-quotient-map. – Ken Mar 11 '24 at 03:46
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Consider a space with three points $a,b,c$ and open sets $\varnothing, \{a\},\{a,b\},\{a,b,c\}$. The projection that identifies $a$ and $c$ is not open (or closed, for that matter.)

Pedro
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  • What does it mean for a projection to identify points? – Muno Sep 21 '17 at 02:38
  • @Muno I mean the map ${a,b,c}\to {0,1}$ that assigns $a$ and $c$ to $1$ and $b$ to $0$, and where ${0,1}$ is given the quotient topology. – Pedro Sep 21 '17 at 14:02
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Let $X$ be a Hausdorff space, and let $a,b$ be a pair of distinct points. Stipulate that $\{b\}$ is not open. Define $f:X\to X\setminus\{b\}$ by $f(z)=z$ if $z\ne b$ and $f(b)=a$. There is a unique topology we may assign the codomain such that $f$ is a quotient map, let's endow that topology on $X\setminus\{b\}$.

Then, pick disjoint neighborhoods $U\ni a,V\ni b$, and note that $f^{-1}(f(U))=U\cup\{b\}$. This set is not open; for, if it were open, then there would be a neighborhood of $b$--call it $W$--contained in $U\cup\{b\}$. So, $\{b\}=(W\cap V)\cap(U\cup\{b\})$ is open, a contradiction.

Karl Kroningfeld
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Partition $X=ℝ$ (under the usual topology) into two equivalence classes $(-\infty, 0]$ and $(0,\infty)$ to get a two-point quotient space $X'$ (called the Sierpiński space) and let $q:X\to X'$ be the canonical projection mapping that takes each element of $ℝ$ to its equivalence class. Then $q$ is a quotient mapping that satisfies $q^{-1}(q((-\infty,0)))=(-\infty,0]$.

mathematrucker
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An open map is: for any open set in the domain, its image is open

A quotient map only requires: for any open preimage in the domain, its image is open

A quotient map may not be an open map because there may exist a set such that it is open but is not a preimage, i.e., the preimage of the image of an open set may not be the open set itself.

Acblade
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